Lucas Oliveira

Modelling a Functional Building

Math Internal Assessment Type 2

As an architect, the task that has been assigned is to design a building with asomewhat elliptical roof structure in which office blocks that should follow certainspecifications- are to be built inside.

Before going on to elaborate a design for this structure, there are somepreviously determined measurements that should be mentioned. To begin with, thewhole building has a rectangular base 150 m long and 72 m wide. In addition, the roofsheight should follow the interval:

36 m g 54 m (where g is the buildings height)

Each room in the office structure has to have at least 2.5 m in height.

The roof would have somewhat of a parabolic shape, making it possible to

generate a model that describes its physical form. When the height of the structure is 36m, and its width is 72 m, we can set the curve on a pair of axis to determine itsintercepts and its opening width. Let us position the ellipse in a way that it issymmetrical in relation to the y-axis (more for convenience purposes). The standardequation for a quadratic function is given by:

y = ax2 + bx + c

However, given that the ellipse is symmetrical to the y-axis, its axis of symmetryis equal to zero. So,

=0

b=0

Since b = 0, the model is no longer described by y = ax2 + bx + c, but by

y=ax2+c. The roofs height in this situation is said to be 36 m high, hence the shapesmaximum point has to have 36 as its y coordinate since it rests at the peak of the ellipse.The maximum points x coordinate is given by the axis of symmetry, which aspreviously found is equal to zero. Hence, the maximum point is (0, 36) and so the c= 36in the model. (0, 36) Lucas Oliveira

In this coordinate system, the x-axis represents ground level (where y = 0), sincepossible underground structures belonging to the roof are not to be considered. Also, thewidth of the roof is 72 m, but since the shape is symmetrical to the y-axis, half of theshape is to the y-axiss left and the other half to its right. Therefore, the y-axis cuts thewidth in half, which means that the shapes roots are given by (-36,0) and (36,0). So, inorder to find the constant a, let us plug in one of the two points found in the equation y= ax2 + 36:

The graph below represents the model found. It also represents a twodimensional view of the roof structure. The red parabola would be the roof itself whilethe area limited by the x-axis and the parabola is the area inside the building. The graphfor this model is the following: Lucas Oliveira

Now, the roof alone does not make up for the whole building. There are otherkey components to it, like a possible cuboid that would hold several offices. In order toget the most out of the roof structure with the height of 36 m, finding the maximumvolume of the cuboid that fits in this structure is required. Let us illustrate this:

x x

Looking at the building in a two dimensional way, the cuboid would fit in theroof structure in a way similar to the one represented above. The variables x and y canassume any value, but since this situation requires the volume to be at its maximum, wehave to determine what values of x and y will allow the area of the rectangle above tooccupy the maximum space underneath the parabola. So to start with, it is necessary torelate both x and y in a single expression. The area of a rectangle is given by the productof its length and width. Ergo,

(x+x)(y)= A (where A is the cuboids area)

2xy = A 1 2 2x ( x + 36) = A 36

1 3 x + 72 x = A 18

Since we want to find the maximum volume of the cuboid, it is necessary to

optimize the situation by finding the derivative of the function given and equating it tozero. When a derivative is equated to zero, the x values give the functions maximumand minimum point. Lucas Oliveira

1 2 A= x +72 6 1 2 0= x +72 6 x2 = 432 x = 12 3 (12 3 20.7846)

These two values for x give the maximum size of the cuboid that fits inside theroof structure. The width goes from the point -12 3 to the point +12 3 , so its totalsize is 24 3 (approximately 41.5692). To find the height of cuboid, the y value, simplyplug-in one of these two points found on the model previously developed: 1 y= (432) + 36 36 y = -12 + 36 y = 24 The length of the cuboid is the same as the length of the building itself (150 m).So, to find the maximum volume of the cuboid that fits inside the roof structure: Length: 150m , width: 24 3 m, height: 24m V = 24 x 24 3 x 150 V 149649.12 m3 Now, the roof structure could have any height value between 36 m g 54 m,suggesting that there might be a change in the cuboids dimensions. The model for the roof structure is given by a broader function: y = ax2 + g (where g is the roofs height and a is the roofs opening) In a sense, by transcribing the function in this manner, one can have morefreedom to decide what values of g one wishes to use. But as there are two unknownvariables (a and g) that are dependent on one another. Hence, there is a way to rewrite ain terms of g. Plugging in an unchangeable point (-36, 0): 0 = a(-36)2 + g -1296a = g g a= 1296 Hence, the broader function could be rewritten as: g 2 y= x +g 1296 Lucas Oliveira

As previously discovered, to find the cuboids maximum volume, let us connect

these equations into one single volume expression. As mentioned earlier, the cuboidswidth is given by (x + x), 150 m is the length of the cuboid and the building itself, andthe height is given by a y value that can be substituted by the equation found above.Thus: g 2 V = 2x (150) ( x + g) 1296 g 2 V = 300x ( x + g) 1296 25g 3 V = 300gx - x 108 To find the maximum volume, it is necessary to optimize the equation above byfinding its derivative and setting it to be equal to zero: 25g 2 V= 300g x 36 25g 2 0 = 300g x 36 25g 2 x = 300g 36 x2 = 432 x = 12 3

Once again, the values found for x are the same, meaning that the width of thelargest possible cuboid, regardless of the structures height, will always be 24 3 m.However, the cuboids height can vary for different values of the roofs height. Using the formula function in Microsoft Excel 2007, it is possible to find thevalues of the cuboids height for different roof heights.*All of the values below are in meters.*periods are represented by commas Roof's Cuboid's Height Height 36 24,0000 38 25,3333 40 26,6667 42 28,0000 44 29,3333 46 30,6667 48 32,0000 50 33,3333 52 34,6667 54 36,0000

*where the width (24 3 m ) and the length (150 m) are the same for all of thedifferent heights Lucas Oliveira

Both the cuboids length and width are kept constant while the cuboids height g 2changes as the roofs height changes. The equation y= x +g shows the 1296relationship between three variables (y: the cuboids height, g: the roofs height, and x:the cuboids width). Since it has been found out that x is constant for any value of g ory, it is plausible to say that y is directly proportional to g. So, in order to find y in termsof g, let us simply substitute the unchangeable value for x, and simplify the expression: g y= (-12 3 )2 + g 1296 g y= +g 3 2g y= 3 Although the cuboid is occupying a large amount of space inside the wholebuilding, there are still areas that are left empty. Therefore, it is possible to calculate theratio of the empty space and the space occupied by the cuboid. To find the volume ofthe whole building (Vb), it is necessary to use calculus integrals: 36 g 2 36 (150) ( 1296 x + g)dx

So, the total volume of the building is given by 7200g and the volume of the cuboid (Vc)is given by the product of its length, width, and height. Therefore, the volume of theempty (Ve) space is given by: Ve = Vb - Vc 2g Ve = 7200g (150 *24 3 * ) 3 Ve = 3043.0781g

Once again, using the formula tool in Microsoft Excel 2007, a ratio can bedetermined for each different structure height value:*All of the values below are in meters.*periods are represented by commas Lucas Oliveira

As the table shows, the ratio between the empty space and the volume occupiedby the cuboid does not change. This means that for any value of the buildings heightwithin its given limitations, the ratio between the empty space and the volume of thelargest cuboid will always remain the same. So, increasing the roofs height willincrease the volume of the largest cuboid, but it will also increase the volume of emptyspace proportionally. The cuboid is the block where offices are going to be set in. Each floor cannot beless than 2.5 m tall, which means that the number of office floors is dependent on thecuboids height. Let us illustrate the situation for a better understanding:

*where red is the whole building, black is the cuboid, and green is the officefloorsAs the illustration above shows, a different number of floors will fit inside the cuboidwith different heights. Not only that but each of these floors will have exactly the sameground area and be at least 2.5 m tall. The buildings height is limited to the interval36m g 54 m, and it was previously found that the buildings height is directionally Lucas Oliveira

2gproportional to the cuboids height by the equation: y = .So, by rearranging the 3 3yequation, g = . Plugging in this value in the roofs height inequality: 2 3y 36 54 2 2 2 36*( ) y54*( ) 3 3 24y36 The new interval for y values found above reveals the minimum and maximumheight for the office block. Now, since each floor has to be at least 2.5 m tall, to find the number of floors acertain cuboid may have simply requires us to divide the height of the office block by2.5. Let us calculate the number of floors for each y value:*periods are represented by commas

*where z is the height of the cuboid

Now, the area of the base of the cuboid will always have a constant value, for thearea of the base can be found by the product of its width and length. As previouslyfound, the width of any cuboid within certain specifications is 24 3 m, and its lengthwill be the same as the buildings length: 150m. Therefore, the area of the base is foundby: Area of base = 150 * 24 3

Area of base = 3600 3 m2

So, for the number of floors in the office block, the amount of floor areaincreases accordingly. The only thing it takes is to multiply the area of the base by thenumber of floors. Therefore, let us calculate the total office floor area for cuboids withdifferent floor numbers:

Number of Total Office

Joining both tables 5 and 6 in a single table shows how the height of the buildingchanges the total office floor area. Number of floors Height of Cuboid Total Office Floor Area 9 24z<25 56118,45 10 25z27,5 62353,83 11 27,5z<30 68589,21 12 30z<32,5 74824,59 13 32,5z<35 81059,98 14 35z<36 87295,36 Lucas Oliveira

So, as the table above demonstrates, when the cuboid is at its maximum height,there are more floors that can be built, which means more total floor area. In a way, thisis fairly simplistic. In all of the calculations and examples so far, it has been considered that thebuildings faade was located at the bases smallest side. However, if the faade is saidto be put on the longest side of the base, some of the structural values could be altered.Let us illustrate the situation:

Let us set a more general equation to begin with.

y=ax2+g In this situation, the full width of the roof structure is 150 m, which implies thatits roots are (-75, 0) and (75, 0). Therefore, by plugging in one of these values in theequation above, a can be found in terms of g. Since the height of the building isunknown at this stage, the model (to avoid confusion let us call it model2) will be keptin terms of the variable g. 0 = a(-75)2 + g g a= 5625 g 2 y= x +g 5625 As previously done, in order to find the cuboid with maximum volume thatwould fit in the structure, it is necessary to optimize the expression that will be formedusing the model2: g 2Area of the curve y: (2x)( x +g) 5625 2g A = 2xg - x3 5625Finding the derivative and equating it to zero: 2g A = 2g - x2 1875 Lucas Oliveira